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COMBINATORICS

by rafodr » Wed Sep 09, 2009 1:30 pm
Q: How many different positive integers having six digits are there, where exactly one of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other two digits is a 7 or an 8?

A)360
B)720
C)840
D)1080
E)1440

Please, I would appreciate some help with this.

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by bharathh » Wed Sep 09, 2009 3:13 pm
Here's my way of doing this.. I'm probably wrong but here goes.

There are 6 digits.

Let's place the 3

There are 6 possible positions it can take... so choose one.

Then the 5 .. Now that you've placed the 3 , there are 5 possible positions to choose from.

Now the 4s. You want to choose 2 out of 4 positions .. so 4C2 = 6 ways

You're left with a 7 and an 8. If you choose to put the 7 in first. There are 2 ways to put it in and then 1 way to put the 8 in... Similarly if you chose to put the 8 in first, there are 2 ways to put the 8 in and then one way to put the 7 in.

so 2*1 * 2*1 = 4

What is the possibility for all these to occur? multiply the choices = 4*6*5*6 = 720 ways to create this number.

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by praky_rules » Wed Sep 09, 2009 3:40 pm
arrange 344577 in 6!/2!2! ways
arrange 344578 in 6!/2! ways
arrange 344588 in 6!/2!2! ways

Note that 6 digit numbers that come from one set will NEVER overlap with the other two.

Sum = 6! = 720 ways.